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Failure mode support of the SN65LVCP22

Daniel Snider
Prodigy 10 points
Other Parts Discussed in Thread: SN65LVCP22

We need to understand the failure mode capabilities of SN65LVCP22.  There are two basic questions.

1-      What failure modes does the part support?

2-   Can the outputs be driven to a ground and be able to recover without being damaged.  Think of durations of hours not seconds.

Thanks

Dan Snider

  • 0
    TI__Guru 62920 points

    Hi Dan,

    Can you please elaborate a bit on what you mean by failure modes?  What sort of failures are you talking about?

    The LVDS outputs will limit their output current to +/- 12 mA when they are both shorted to ground, and can stay in this mode continuously without damage to the device.

    Regards,
    Max Robertson
    Analog Applications Engineer
    m-robertson@ti.com

     

  • 0 in reply to Miguel Robertson
    TI__Guru 62920 points

    This was answered offline a couple of weeks ago, but for the benefit of the E2E community here is how the SN65LVCP22 handles various faults:

    • Inputs shorted together – This is not an issue for the LVCP22.  It just means that the input differential signal will be reduced to 0 V.  The resulting output will be indeterminate since there is no differential input to detect, but no damage should occur to the device.
    • Inputs shorted to ground – This is also OK.  The inputs are high impedance so they will not source appreciable current (the maximum spec is 10 uA).  The voltage on these pins can go as low as -0.7 V before the diode protection kicks in.  As long as you are above -0.7 V (and below 4.3 V) no damage will result.
    • Outputs shorted to ground – The current will be limited to +/- 12 mA.  The power dissipation resulting from this is too low to cause any thermal issues, so runaway should not be a concern.
    • Outputs shorted together – The LVCP22 has a current-mode output, meaning that it establishes an output voltage by steering current through a termination resistance.  Shorting the output just makes this resistance 0 Ohms.  The same amount of current will flow from one pin to the other as in normal operation, it just won’t establish any differential voltage.  No damage will result.

    Best regards,
    Max